Question

If the threshold wavelength (${\lambda }_0$) for ejection of an electron from metal is 330 nm, then work function for the photoelectric emission is $6\times {10}^{16+y}$. The value of $y$ is:

Answer 1

As we know,
$W=h{v}_0=\frac{hc}{{\lambda }_0}=\frac{6.6\times {10}^{-34}Js\times 3\times {10}^{8}m}{330\times {10}^{-9}m}=6\times {10}^{-19}$.